Find the area of the yellow shaded area govern the larger pentagon side s=5.5 and area of the smaller pentagon (green) A_(pentagon) = 7.59 cm^2?

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##A_Delta = 1/2 bxxh ~~ 5/ 2 (2.1xx3.23) ~~17##

Given: The verge of the of the larger recognized pentagon, ##s=5.5## Area meaner pentagon, ##A_(pentagon) = 7.59##

Required: Area of the yellow retreating triangles of the pentagram?

Solution Strategy: 1) Find ##theta and alpha## intelligent we keep a recognized pentagon ##alpha = 180[(n-2)]/n## n is calculate of verges, ##n=5## ##alpha=180(3)/5=108^0, :. theta=72^0## 2) Find the verge of the mean pentagon intelligent that, ##A_("mean pentagon")=7.59## ##A_("mean pentagon") = 1/4 sqrt(5(5+2sqrt(5) )) a^2## ##a = 25^(3/4) sqrt(A)/(5(sqrt20+5)^(1/4)) =25^(3/4) sqrt(7.59)/(5(sqrt20+5)^(1/4)) ~~ 2.1## 3) Find the altitude/height of the yellow triangle: ##tan72= (Opp)/(Adj)= h/(b/2); h=1.05tan(72) ~~ 3.23## 4) Area of the triangle and increase by 5 ##A_Delta = 1/2 bxxh ~~ 5/ 2 (2.1xx3.23) ~~17##

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